Set Theory Practice Solution5

In this page set theory practice solution5 we are going to see
solution of practice questions from the worksheet set theory practice
questions2.

Question 4:

verify n (AUBUC)=n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n (A∩C)+n(A∩B
∩C)

(i) A = {4,5,6},B = {5,6,7,8} and C = {6,7,8,9}

Solution:

To verify the condition n (AUBUC)=n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n (A∩C)+n(A∩ B
∩C) we have to find number of terms in A,B,C number of terms in A U B ,B U C and C U A and also number of terms in (A∩ B
∩C)

A = {4,5,6}

B = {5,6,7,8}

C = {6,7,8,9}

n (A) = 3 n (B) = 4 n (C) = 4

(A ∩ B) = {5,6}

n (A ∩ B) = 2

(B ∩ C) = {6,7,8}

n (B ∩ C) = 3

(A ∩ C) ={6}

n (A ∩ C) = 1

(A ∩ B
∩ C) = {6}

n (A ∩ B
∩ C) =1

n (A U B U C)=n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n (A∩C)+n(A ∩ B
∩ C)

= 3 + 4 + 4 - 2 - 3 - 1 + 1

= 11 - 6 + 1

= 12 - 6

= 6

(ii) A = {a,b,c,d,e} B = {x,y,z} and C = {a,e,x}

Solution:

A = {a,b,c,d,e} B = {x,y,z} and C = {a,e,x}

n (A) = 5 n (B) = 3 n (C) = 3

n (A ∩ B) = 0

B ∩ C = {x}

n (B ∩ C) = 1

C ∩ A = {a,e}

n (C ∩ A) = 2

n (A ∩ B
∩ C) = 0

n (A U B U C)=n(A)+n(B)+n(C)-n(A∩B)-n(B∩C)-n (A∩C) + n(A ∩ B
∩ C)

= 5 + 3 + 3 - 0 - 1 - 2 + 0

= 11 - 3

n (A U B U C) = 8

Question 5:

In a college,60 students enrolled in chemistry,40 in physics,30 in
biology,15 in chemistry and physics,10 in physics and biology,5 in
biology and chemistry. No one enrolled in all the three. Find how many
are enrolled in at least one of the subjects.

Solution:

Let A,B and C are the sets enrolled in the subjects Chemistry,Physics and Biology respectively.